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An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.

iren2701 [21]

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

a) $\vec{v}$ = 70·i

b) The package will reach ground in approximately 12.77 seconds.

c) The speed of the package as it lands is approximately 145.51 m/s.

d) The path of the package based on a stationary frame of reference is parabolic

e) The path of the package as seen from the plane is directly vertical downwards

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, $\vec{v}$ = 70·i

b) The time it will take the package to reach the ground, t, is given by the formula;

$\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}$

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

$\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}$

Which gives;

$\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}$

The time it will take the package to reach the ground, t ≈ 12.77 seconds

c) The vertical velocity just before the package reaches the ground, $v_y$ , is given as follows;

$v_y^2$ = 2·g·h

Therefore;

$v_y$ = √(2·g·h)

Which gives;

$v_y$ = √(2 × 9.81 × 800) ≈ 125.28

$v_y$ ≈ 125.28 m/s

Which gives; $\vec{v}$ = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ 143.51 m/s

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a parabolic shape.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating directly vertical

downwards.

Learn more about projectile motion here:

brainly.com/question/1130127

6 0

2 years ago

Please help... !!!!!!!

Scilla [17]

Answer:

1) C. The precise location of weather-related phenomena

2) C. It shows that the galaxies are moving farther away.

Explanation:

1) The global positioning system can be used in many ways, like in navigation, mapping and surveying the land. It is also used to observe the small changes taking place in the movement of objects on earth. They are usually used to study earthquakes.

2) When light from the galaxies is shifted towards the longer wavelengths, it is called red shift. When light shifts towards the shorter wavelengths, it is called blue shift. Big bang theory states how the primordial universe started expanding from a hot and dense singularity, in the process of evolution of the universe.

The universe is expanding always. Galaxies that are a part of the universe keep expanding and move farther away from each other. The galaxies that are very far off appear to move much faster than the nearby galaxies.

6 0

3 years ago

What is used as evidence for sea-floor spreading?

raketka [301]

Answer:

Sea-floor spreading occurs in the oceanic ridges. In there, volcanic activity, together with the gradual movement of the bottom, form new oceanic crust. This allows a better understanding of the continental drift explained by the theory of plate tectonics.

The greatest evidence for Sea-floor spreading is the oceanic trenches, the oceanic ridges, the magma protruding to the surface and the new seafloor.

In previous theories, continents were assumed to be transported across the sea. Harry Hess, in the 1960s, proposed the idea that the seabed itself moves as it expands from a central point. The theory is now accepted, and the phenomenon is thought to be caused by convection currents in the upper layer of the mantle.

4 0

4 years ago

Read 2 more answers

A 2200 kg SUV traveling at 23.9 m/s can be stopped in 0.16 s if it hits a concrete wall. Assume that a 60 kg person was in the c

Jlenok [28]

Answer:

8962.5 N

Explanation:

From fundamental kinematics equations, acceleration, a is given by $a=\frac {v-u}{t}$ where v is final velocity, u is initial velocity and t is time

We also know that force is a product of mass and acceleration of an object and substituting $a=\frac {v-u}{t}$ in the formula F=ma we obtain

$F=ma=m\frac {v-u}{t}$ where m is the mass of the body

$F=60\times\frac {0-23.9}{0.16}=-8962.5 N$

Therefore, the force is 8962.5 N

4 0

4 years ago

You’re driving your car towards an intersection. A Porsche is stopped at the red light. You’re traveling at 36 km/h (10 m/s). As

34kurt

Your position at time $t$ , relative to the stop line:

$x_1=-15\,\mathrm m+\left(10\dfrac{\rm m}{\rm s}\right)t$

The Porsche's position:

$x_2=\dfrac12\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2$

a. You pass the Porsche immediately after the time it takes for $x_1=x_2$ :

$-15\,\mathrm m+\left(10\dfrac{\rm m}{\rm s}\right)t=\dfrac12\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=2.3\,\rm s$

at which point you both will have traveled 7.8 m from the stop line.

b. The equation in part (a) has two solutions. The Porsche passes you at the second solution of about $t=4.4\,\rm s$ , at which point you both will have traveled 29 m.

c. At time $t$ , the Porsche is moving at velocity

$v=\left(3\dfrac{\rm m}{\mathrm s^2}\right)t$

so that at the moment it passes you, its speed is 13 m/s, which is about 46.8 km/h and below the speed limit, so neither of you will be pulled over.

6 0

4 years ago